5c3 is because of the 4's and 7's in rows 1 and 2 (ie Box 7 and Box 9 both contain 4's and 7's in rows 1 and 2 --> therefore box 8 must contain either a 4 or 7 in both r3cd and r3cf and there's already a 8 in r3ce --> hence there must be a 5 in row 1 of box . Couple the 5 in row 1 of Box 8 along with the 5 in r2cg of Box 9, that means the only place for the 5 in Box 7 is r3cc.

It would appear that SamGJ is now compiling "Very Hard"
puzzles that are not amenable to full solution by "Mandatory
Pairs" methods - unlike a few weeks ago when the 'v.hard'
were often easier than the plain 'hard'.

This one was solved after "normalising" column 8 and row 7.

"Normalising" is the splitting of the candidate string for a row
or column into two or (rarely) three subgroups where each
subgroup portrays all possible candidates for a number of
cells equal to the number of candidates in the subgroup.

An example would be 26,378,237,26,38.
The candidate string for the line is 23678.
This MUST have five digits in it as five cells are involved.
Normalisation would spot the pair of 26 items
This gives (26)(2378) as subgroups BUT the second group has
a repeat of one of the digits in the first group - a real trigger
to paying attention!
Removing the '2' from the 237 cell gives the normalised set
of (26)(378). The second group now has three members (with
values 378,37,38) and covers three cells.

The original string 23678 is "congruent" for the line (as it has
five distinct digits and covers five cells) but it is not "normalised"
as it is capable of division into two congruent substrings (albeit
by - beneficial - modification of one of the cell profiles).